RMS speed in kinetic energy equation for gas

[zanyzoya]

I have been pondering. Why is it that we use the rms speed in the equation Ek = 1/2 m v_rms², as opposed to just the mean speed²

 

[mjc123]

Because kinetic energy is 1/2 mv². The kinetic energy of molecule i is 1/2 mv_i², so the average kinetic energy is 1/2m * (average of v²). If you used the square of the mean speed you would get a different, wrong answer. (The mean velocity is, of course, zero.)

 

[lychette]

Because kinetic energy is 1/2 mv². The kinetic energy of molecule i is 1/2 mv_i², so the average kinetic energy is 1/2m * (average of v²). If you used the square of the mean speed you would get a different, wrong answer. (The mean velocity is, of course, zero.)

I agree

 

[zanyzoya]

Thanks for your help with that mjc 123 and lychette, it makes much more sense now.

 

[Shreyans Jain]

We are using V(rms) as mean velocity of the system will come out to be zero as particles are moving randomly in all the directions, So Rms (Root Mean Square) is taken.V^2(Rms) = N1 x V1^2 + N2 x V2^2 Divided by N1 + N2Here N is molecule and V is its velocity. As it is a square all velocities will become positive.

 

[mjc123]

That is true, but as pointed out above it is not the only reason. You could use the mean speed (magnitude of the velocity), but that would give the wrong answer because KE is proportional to v², so you need the average value of v². The mean square is not equal to the square of the mean.